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Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
A trapezium of parellel sides 10cm and 21cm and height 8cm is inscribed in a circle of radius 7cm. calculate the area of the region not covered by the trapezium.
π =\(\frac{22}{7}\)
A
84cm\(^2\)
B
80cm\(^2\)
C
30cm\(^2\)
D
94cm\(^2\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of circle = \pir\(^2\) = \(\frac{22}{7}\) * 7\(^2\) = 154cm\(^2\)
Area o Trapezium = \(\frac{(a + b) h}{2}\) = \(\frac{(10 + 21)8}{2}\) = 124cm\(^2\)
Area of the region not covered = Area of (Circle - Trapezium)
= 154 - 124
= 30cm\(^2\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61356)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
find the value of (x+y)
Attached Figure / Diagram
A
215°
B
70°
C
135°
D
145°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
sum of angle in a straight line = 180^\circ
35 + m = 180
m = 180 - 35 = 145^\circ
Corresponding angles are congruent i.e equal to each other
Angle X and M are congruent = 145
sum of angle in a straight line = 180^\circ
110 + n = 180
m = 180 - 110 = 70^\circ
Angle Y and N are congruent = 70^\circ
\(\therefore\) ( x + y) = ( 145 + 70)^\circ
(x + y) = 215^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61422)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t.
Attached Figure / Diagram
A
130°
B
115°
C
58°
D
68°
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Alternate segments are equal : R^\circ = 64^\circ
An isosceles triangle has two angles ( Q^\circ & N^\circ ) equal as A^\circ
R^\circ + A^\circ + A^\circ = 180^\circ
2A^\circ= 180 - 64
2A^\circ = 116^\circ
A = \(\frac{116}{2}\)
A = 58.
: Q^\circ = 58, N^\circ = 58 and T^\circ = 58
As Alternate segments are equal: T = Q
Source: WAEC / JAMB / NECO Classroom Archive (Record #61423)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a circle. find the value of x.
Attached Figure / Diagram
A
50°
B
30°
C
80°
D
100°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Opp. angles in a cyclic quadrilateral always add up to 180^\circ
∠P + ∠R & ∠Q + ∠S = 180
x + x+y = 180
2x + y = 180... i
2y - 30 + x = 180
2y + x = 180 + 30
x + 2y = 210 ... ii
Elimination method:
(2x + y = 180) * 1 --> 2x + y = 180
(x + 2y = 210) * 2 --> 2x + 4y = 420
Subtracting both equations
- 3y = - 240
y = 80^\circ
using eqn i
2x + y = 180
2x + 80 = 180
2x = 100
x = 50^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61425)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
Given that sin x = \(\frac{3}{5}\), 0 ≤ x ≤ 90, evaluate (tan x + 2cosx)
A
2\(\frac{11}{20}\)
B
2\(\frac{7}{20}\)
C
\(\frac{11}{20}\)
D
\(\frac{1}{20}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Sin x = \(\frac{opp}{hyp}\)
sinx = \(\frac{3}{5}\)
using Pythagoras' theorem
hyp\(^2\) = opp\(^2\) + adj\(^2\)
adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9
adj\(^2\) = 16
adj = \sqrt 16
adj = 4.
tanx = \(\frac{opp}{adj}\)
= \(\frac{3}{4}\)
cosx = \(\frac{adj}{hyp}\)
= \(\frac{4}{5}\)
(tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\))
= \(\frac{15 + 32}{20}\)
= \(\frac{47}{20}\) or
2 \(\frac{7}{20}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61427)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.
If angle ABC equals 158°, find ∠ADE
Attached Figure / Diagram
A
112
B
90
C
68
D
22
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the given diagram;
Source: WAEC / JAMB / NECO Classroom Archive (Record #61429)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a circle. \(\overline{|PQ|}\) = \(\overline{|QS|}\). ∠SPR = 26° and the interior angles of PQS are in the ratio 2:3 :3.
Calculate: (i) PQR; (ii) RPQ; (iii) PRQ
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where X is a real number.
If |PQ| = 29units, find the value of x.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) ∠PSQ = ∠QPS
∠PQS = \(\frac{2}{8}\) x 180^\circ
=45^\circ
∠SPQ = ∠PSQ
= \(\frac{3}{8}\) x 180 = 67.5^\circ
∠SQR = ∠SPR= 26^\circ
∠PQR = ∠PQS + ∠SQR
= 45^\circ+ 26^\circ
= 71^\circ
(ii) ∠RPQ = ∠SPQ - ∠SPR
= 67.5^\circ- 26^\circ
= 41.5^\circ
(iii) ∠PRQ = 180^\circ - 41.5^\circ - 71^\circ
= 67.5^\circ
(b) |PQ| = \(\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}\)
29 = \(\sqrt{ (5 - 7)^2 + (x - 3)^2}\)
Take the square of both sides
\(29^2\) = \(-2)^2 + (x - 3)^2
841 = 4 + x\(^2\) - 6x + 9
841 = x\(^2\) - 6x + 13
0 = x\(^2\) - 6x + 13 - 841
x\(^2\) - 6x - 828 = 0
Using the formula method
where a = 1, b = -6, and c = -828
x = 31.93095 or - 25.93095
Therefore x ≈ 31. 93 or x = - 25.93
Source: WAEC / JAMB / NECO Classroom Archive (Record #61563)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
(a) On Sam's first birthday celebration, his grandfather deposited an amount of S 1,000.00 in a bank compound at 4 % interest annually.
Find how much is in the account if Sam is 4 years old.
(b) In the diagram above, ABCD are points on the circle centre O. If |AB| = |BC| and ∠ADC= 50°, find ∠BAD.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Amount at end of the 2nd year = \(\frac{104}{100}\) x $1000
= $1,040.00
Amount at end of 3rd year = \(\frac{104}{100}\) x $1040
= $1,081.60
Amount at end of 4th year = \(\frac{104}{100}\) x $1081.6
= $1,124.86
(b) ∠ADC + ∠DCA+ ∠CAD = 180^\circ
50^\circ + 90^\circ+ ∠CAD = 180^\circ
∠CAD = (180 - 140)^\circ
= 40
∠ADC + ∠ABC = 180^\circ
50^\circ + ∠ABC = 180^\circ
∠ABC =130^\circ
∠BAC+ ∠BCA + ∠ABC = 180^\circ
But ∠BAC = ∠BCA
2∠BAC+ ∠130^\circ = 180^\circ
2 ∠BAC=50^\circ
∠BAC = 25^\circ
∠BAD = ∠CAD + ∠BAC
= 40^\circ+25^\circ = 65^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61564)
Mathematics
WAEC / WASSCE 2021
Number Bases
Solution
Simplify: (11\(_{two}\))\(^2\)
A
1001\(_2\)
B
1101\(_2\)
C
101\(_2\)
D
10001\(_2\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))
= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\)
= \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\)
= \({2 + 1} \times ({2 + 1})\)
= 3 \(\times\) 3
= 9\(_{10}\) or
1001\(_2\) from
2
9
2
4 r1
2
2 r0
2
1 r0
0 r1
Source: WAEC / JAMB / NECO Classroom Archive (Record #61138)
Mathematics
UTME / JAMB 2021
Indices
Solution
Correct 241.34(3 x 10\(^{-3}\))\(^2\) to 4 significant figures
A
0.0014
B
0.001448
C
0.0022
D
0.002172
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)
(3 x 10-\(^3\))\(^2\)............(B)
= 3\(^2\)x\(^2\)
= \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\)
(Note that x\(^2\) = \(\frac{1}{x^3}\))
= 24.34 x 3\(^2\) x \(\frac{1}{10^6}\)
= \(\frac{2172.06}{10^6}\)
= 0.00217206
= 0.002172(4 s.f)
Source: WAEC / JAMB / NECO Classroom Archive (Record #60670)
Mathematics
UTME / JAMB 2021
Indices
Solution
Find the mean deviation of 1, 2, 3 and 4
A
1.0
B
1.5
C
2.0
D
2.5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
_
Mean deviation = Σ|x - x|
n
_
x = 2.5
= |1 - 2.5| + |2 - 2.5| + |3 - 2.5| + |4 - 2.5|
4
= 4/4 = 1
Source: WAEC / JAMB / NECO Classroom Archive (Record #60671)
Mathematics
UTME / JAMB 2021
Indices
Solution
Simplify and express in standard form \(\frac{0.00275 \times 0.00640}{0.025 \times 0.08}\)
A
8.8 x 10-1
B
8.8 x 10-2
C
8.8 x 10-3
D
8.8 x 103
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\)
Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\)
= \(\frac{88}{10^4}\)
88 x 10-4 = 88 x 10-1 x 10-4
= 8.8 x 10-3
Source: WAEC / JAMB / NECO Classroom Archive (Record #60677)
Mathematics
WAEC / WASSCE 2021
Indices
Solution
Solve: 2\(^{\sqrt{2x + 1}}\) = 32
A
13
B
24
C
12
D
11
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2\(^{\sqrt{2x + 1}}\) = 32
2\(^{\sqrt{2x + 1}}\) = 2\(^5\)
\(\sqrt{2x + 1}\) = 5
square both sides
2x + 1 = 5\(^2\)
2x + 1 = 25
2x = 25 - 1
2x = 24
x = \(\frac{24}{2}\)
x = 12
Source: WAEC / JAMB / NECO Classroom Archive (Record #61139)
Mathematics
WAEC / WASSCE 2021
Indices
Solution
If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.
A
-4
B
4
C
1
D
-1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\)
= 2\(^4\) * 2\(^{(x + 1)}\) = 2\(^{2x}\) * 2\(^{3(1 - x)}\)
--> 4 + x + 1 = 2x + 3 - 3x
collect like terms
--> x - 2x + 3x = 3 - 1 - 4
--> 2x = -2
--> x = -1
Source: WAEC / JAMB / NECO Classroom Archive (Record #61153)
Mathematics
UTME / JAMB 2021
Logarithms
Solution
Simplify 2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9
A
1 - 4 log3
B
-1 + 2 log 3
C
-1 + 5 log2
D
1 - 2log 2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9
[\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\)
= log \(\frac{72}{125}\)
= log \(\frac{5}{2}\)
= log \(\frac{10}{4}\)
= log 10 - log 4
= log10 - log2\(^2\)
= 1 - 2 log2
Source: WAEC / JAMB / NECO Classroom Archive (Record #60710)